Question

A first order reaction is 12.5% completed in 20 minutes at 300k.When same reaction is carried out at 315k, it is 12.5% complete in 2.5 minutes, what is the activation energy of reaction (in k.Cal/mol) ?

Answer 1

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Answer 2

Answer:

11.30 kCal/mol is the activation energy of reaction.

Explanation:

$k=\frac{2.303}{t}\log \frac{A_o}{A}$

When reaction is carried reaction is 12.5% completed in 20 minutes at 300 K.

$A_o=x$

A=87.5% of x = 0.875x

t = 20 min

$k_1=\frac{2.303}{t}\log \frac{A_o}{A}$

$k_1=\frac{2.303}{20 min}\log \frac{x}{0.875x}$

$k_1=0.006678 min^{-1}$

When reaction is carried reaction is 12.5% completed in 2.5 minutes at 315 K.

$A_o'=y$

A'=87.5% of x = 0.875x

t' = 20 min

$k_2=\frac{2.303}{t'}\log \frac{A_o'}{A'}$

$k_2=\frac{2.303}{2.5 min}\log \frac{y}{0.875 y}$

$k_2=0.05342 min^{-1}$

Activation energy of reaction:

$\log \frac{k_2}{k_1}=\frac{E_a}{2.303R}[\frac{1}{T_1}-\frac{1}{T_2}]$

$T_1=300K, T_2=315 K$

R = 8.314 J/mol K

$\log \frac{0.05342 min^{-1}}{0.006678 min^{-1}}=\frac{E_a}{2.303\times 8.314 J/mol K}[\frac{1}{300 K}-\frac{1}{315 K}]$

$E_a=\frac{0.90306\times 300 K\times 315 K\times 8.314 J/mol K}{15 K}$

$E_a=47,300.525 J/mol=\frac{47,300.525 }{4184 } kCal/mol$

(4184 J = 1 kCal)

$E_a=11.30 kCal/mol$

11.30 kCal/mol is the activation energy of reaction.